Respuesta :
The initial horizontal velocity of the soccer ball is 16.5 m/s
Explanation:
When we throw a ball, there is a constant velocity horizontal motion and there is an accelerated vertical motion. These components act independently of each other. Horizontal motion is constant velocity motion.
[tex]v_{x f}=v_{x i}=v_{x}[/tex]
[tex]a_{x}=0[/tex], so [tex]x=v_{i x} t+\left(\frac{1}{2}\right) a_{x} t^{2}[/tex] for horizontal motion
[tex]y=v_{i y} t+\left(\frac{1}{2}\right) a_{y} t^{2}[/tex] for vertical motion
Given:
x = 35 m
[tex]a_{x}=0 \mathrm{m} / \mathrm{s}^{2}[/tex]
Need to find [tex]v_{i x}[/tex]
y = - 22 m
[tex]v_{i y}=0 \mathrm{m} / \mathrm{s}[/tex]
[tex]a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}[/tex] (negative sign indicates downward motion)
By substituting all known values, we can solve for 't' value as below
[tex]y=v_{i y} t+\left(\frac{1}{2}\right) a_{y} t^{2}[/tex]
[tex]-22=0(t)+\left(0.5 \times-9.8 \times t^{2}\right)[/tex]
[tex]t^{2}=\frac{-22}{0.5 \times-9.8}=\frac{-22}{-4.9}=4.4897[/tex]
Taking square root, we get t = 2.12 seconds
Now, substitute these to find initial horizontal velocity
[tex]x=v_{i x} t+\left(\frac{1}{2}\right) a_{x} t^{2}[/tex]
[tex]35=v_{i x}(2.12)+\left(0.5 \times 0 \times(2.12)^{2}\right)[/tex]
[tex]35=v_{i x}(2.12)+0[/tex]
[tex]v_{i x}=\frac{35}{2.12}=16.5 \mathrm{m} / \mathrm{s}[/tex]
